(x^8y^2+x^2y^8)/(x^2+y^2)=?

=>

(x^8y^2+x^2y^8)/(x^2+y^2)= x^2y^2(x^6+y^6)/(x^2+y^2)= x^2y^2(x^2+y^2)(x^4-x^2y^2+y^4)/(x^2+y^2)= x^2y^2 (x^4-x^2y^2+y^4), using the identity a^3+b^3 = (a+b)(a^2-ab+b^2).

Therefore, the answer is A.
Answer: A

Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

(x^8y^2+x^2y^8)/(x^2+y^2)=?

A. x^2y^2(x^4-x^2y^2+y^4)
B. x^2y(x^4+x^2y^2+y^4)
C. x^2y^2(x^2+y^2)
D. x^2y^2(x+y)
E. x^4y^4(x^2-y^2)

Let´s explore a particular case and proceed with the "filtering" (i.e., refuting alternatives that do not match the target)!
$$? = {{{x^8}{y^2} + {x^2}{y^8}} \over {{x^2} + {y^2}}}\,\,$$
$${\left. {{{{x^8}{y^2} + {x^2}{y^8}} \over {{x^2} + {y^2}}}\,\,} \right|_{\,\left( {x,y} \right) = \left( {1,1} \right)}}\, = \,\,\,{{1 + 1} \over {1 + 1}}\,\,\, = 1\,\,\,\,\,\,\,\left( {{\rm{TARGET}}} \right)$$
$$\left. \matrix{
\left( {\rm{A}} \right)\,\,\,1 \cdot \left( {1 - 1 + 1} \right) = 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{survivor}}\,\, \hfill \cr
\left( {\rm{B}} \right)\,\,\,1 \cdot \left( {1 + 1 + 1} \right) \ne 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{refuted}} \hfill \cr
\left( {\rm{C}} \right)\,\,\,1 \cdot \left( {1 + 1} \right) \ne 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{refuted}} \hfill \cr
\left( {\rm{D}} \right)\,\,\,1 \cdot \left( {1 + 1} \right) \ne 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{refuted}} \hfill \cr
\left( {\rm{E}} \right)\,\,\,1 \cdot \left( {1 - 1} \right) \ne 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{refuted}} \hfill \cr} \right\}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {\rm{A}} \right)$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.